Answer:
Sorry cant find the answer but i hope you got it right and if you didn't you'll still do great. :)
Step-by-step explanation:
Answer:
To complete the problem statement it is needed:
1.- the volume and weight capacity of the truck, because these will become the constraints.
2.- In order to formulate the objective function we need to have an expression like this:
" How many of each type of crated cargo should the company shipped to maximize profit".
Solution:
z(max) = 175 $
x = 1
y = 1
Assuming a weight constraint 700 pounds and
volume constraint 150 ft³ we can formulate an integer linear programming problem ( I don´t know if with that constraints such formulation will be feasible, but that is another thing)
Step-by-step explanation:
crated cargo A (x) volume 50 ft³ weigh 200 pounds
crated cargo B (y) volume 10 ft³ weigh 360 pounds
Constraints: Volume 150 ft³
50*x + 10*y ≤ 150
Weight contraint: 700 pounds
200*x + 360*y ≤ 700
general constraints
x ≥ 0 y ≥ 0 both integers
Final formulation:
Objective function:
z = 75*x + 100*y to maximize
Subject to:
50*x + 10*y ≤ 150
200*x + 360*y ≤ 700
x ≥ 0 y ≥ 0 integers
After 4 iterations with the on-line solver the solution
z(max) = 175 $
x = 1
y = 1
Answer:
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Step-by-step explanation:
Answer:
Start to end
1) N
2)P
3)J
4)B
5)A
6)i
7)L
8)E
9)G - 14b
10)H
11)O
12)M
13)C
14)D
15)K
16)F
Hope this helps, I've matched it up with the letters starting at the top box first, then going down.
Answer: 3y-y+6=2y+6
Step-by-step explanation:
3y-y =2y and that leaves only 6 to add on . And your answer is 2y+6
